Решите уравнения: б) 2cosx^2-cosx-1 = 0 в) sin^2x+sqrt3*sinXcosX=0

B) cosx=a

2a²-a-1=0

D=1+8=9

a1 = (1-3) / 4=-1/2⇒cosx=-1/2⇒x=π-π/3+2πn=2π/3+2πn U x=4π/3*2πn

a2 = (1+3) / 4=1⇒cosx=2πn

b) Разделим на cos²x≠0

tg²x+√3tgx=0

tgx (tgx+√3) = 0

tgx=0⇒x=πn

tgx=-√3⇒x=-π/3+πn